Chemistry Q59 – JEE-MAIN 2025

Home Chemistry Coordination CompoundsQuestion #346

'X' is the number of electrons in

t_{2 g}

orbitals of the most stable complex ion among

[F e (N H_{3})_{6}]^{3 +}

[F e C l_{6}]^{3 -}

[F e (C_{2} O_{4})_{3}]^{3 -}

and

[F e (H_{2} O)_{6}]^{3 +}

. The nature of oxide of vanadium of the type

V_{2} O_{X}

is :

Clue 1 of 3: Stability of Coordination Complexes

The stability of coordination complexes is influenced by factors like ligand field strength and crystal field stabilization energy (CFSE).

🥷

Ninja Strategy•Recall and Trend Analysis

Recognize that $V_{2} O_{5}$ is a high oxidation state oxide of a transition metal, making it unlikely to be neutral or purely basic, narrowing down to acidic or amphoteric.

🧠 Full Solution Path

Step 1: Determine Fe oxidation state and electronic configuration✦ Active

In all given complexes, the central metal ion is $F e^{3 +}$ . The electronic configuration of $F e^{3 +}$ is $[A r] 3 d^{5}$ .

Step 2: Identify spin state,

t_{2 g}

electrons, and the most stable complex○ Expand

The spectrochemical series is $C l^{-} < H_{2} O < C_{2} O_{4}^{2 -} < N H_{3}$ . For $F e^{3 +}$ ( $d^{5}$ ):

- $[F e C l_{6}]^{3 -}$ ( $C l^{-}$ is weak field): High spin, $(t_{2 g})^{3} (e_{g})^{2}$ . CFSE = 0.

- $[F e (H_{2} O)_{6}]^{3 +}$ ( $H_{2} O$ is weak field): High spin, $(t_{2 g})^{3} (e_{g})^{2}$ . CFSE = 0.

- $[F e (C_{2} O_{4})_{3}]^{3 -}$ ( $C_{2} O_{4}^{2 -}$ is intermediate/strong field): Low spin, $(t_{2 g})^{5} (e_{g})^{0}$ . CFSE = $- 2.0 Δ_{o} + 2 P$ .

- $[F e (N H_{3})_{6}]^{3 +}$ ( $N H_{3}$ is strong field): Low spin, $(t_{2 g})^{5} (e_{g})^{0}$ . CFSE = $- 2.0 Δ_{o} + 2 P$ .

The most stable complex will have the largest negative CFSE. Since $N H_{3}$ is the strongest field ligand, it will induce the largest $Δ_{o}$ . Assuming $Δ_{o} > P$ for $N H_{3}$ with $F e^{3 +}$ , $[F e (N H_{3})_{6}]^{3 +}$ will be the most stable low spin complex. For this complex, the number of electrons in $t_{2 g}$ orbitals is $X = 5$ .

💡 Teacher's Secret Hint

Remember that for a low spin complex to be more stable than a high spin one, the crystal field splitting energy ( $Δ_{o}$ ) must be greater than the pairing energy (P).

Step 3: Determine the nature of the oxide

V_{2} O_{X}

○ Expand

With $X = 5$ , the oxide is $V_{2} O_{5}$ . In $V_{2} O_{5}$ , vanadium is in the $+ 5$ oxidation state. Oxides of transition metals in higher oxidation states tend to be acidic or amphoteric. $V_{2} O_{5}$ is known to be amphoteric, reacting with both strong acids and strong bases.

💡 Teacher's Secret Hint

Recall the general trend: as the oxidation state of a metal increases, the acidic character of its oxide increases. Lower oxidation states are basic, intermediate are amphoteric, and higher are acidic.

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Chemistry Question 59 – JEE-MAIN 2025

Clue 1 of 3: Stability of Coordination Complexes

🧠 Full Solution Path